Rank of a 4x4 Matrix A: Linear Algebra Homework Solution

underacheiver
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Homework Statement


Find the rank of A =
{[1 0 2 0]
[4 0 3 0]
[5 0 -1 0]
[2 -3 1 1]}

Homework Equations





The Attempt at a Solution


i row reduced A to be:
{[1 0 0 0]
[0 1 0 -1/3]
[0 0 1 0]}

where do i go from here?
 
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I think you omitted a last row of zeros. Ok, what does rank mean?
 
The column rank of a matrix A is the maximal number of linearly independent columns of A. Likewise, the row rank is the maximal number of linearly independent rows of A.

Since the column rank and the row rank are always equal, they are simply called the rank of A.
 
underacheiver said:
The column rank of a matrix A is the maximal number of linearly independent columns of A. Likewise, the row rank is the maximal number of linearly independent rows of A.

Since the column rank and the row rank are always equal, they are simply called the rank of A.

Good! So how many linearly independent rows are there? If you have no idea, quote the definition of linear independence.
 
so is it 3? because the 2nd and 4th columns are dependent.
 
underacheiver said:
so is it 3? because the 2nd and 4th columns are dependent.

Yes, the second and fourth columns being dependent means the rank is at most 3. Now you have to check that the three remaining vectors are linearly independent. It's easier to see this if you look at the row reduction.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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